Mongoose limit/offset and count query

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Solution 1: [1]

I suggest you to use 2 queries:

  1. db.collection.count() will return total number of items. This value is stored somewhere in Mongo and it is not calculated.

  2. db.collection.find().skip(20).limit(10) here I assume you could use a sort by some field, so do not forget to add an index on this field. This query will be fast too.

I think that you shouldn't query all items and than perform skip and take, cause later when you have big data you will have problems with data transferring and processing.

Solution 2: [2]

Instead of using 2 separate queries, you can use aggregate() in a single query:

Aggregate "$facet" can be fetch more quickly, the Total Count and the Data with skip & limit


      //{$sort: {...}}



        "stage1" : [ {"$group": {_id:null, count:{$sum:1}}} ],

        "stage2" : [ { "$skip": 0}, {"$limit": 2} ]
     {$unwind: "$stage1"},
      //output projection
        count: "$stage1.count",
        data: "$stage2"


output as follows:-

     count: 50,
     data: [

Also, have a look at

Solution 3: [3]

    { '$match'    : { } },
    { '$sort'     : { '_id' : -1 } },
    { '$facet'    : {
        metadata: [ { $count: "total" } ],
        data: [ { $skip: 1 }, { $limit: 10 },{ '$project' : {"_id":0} } ] // add projection here wish you re-shape the docs
    } }
] )

Instead of using two queries to find the total count and skip the matched record.
$facet is the best and optimized way.

  1. Match the record
  2. Find total_count
  3. skip the record
  4. And also can reshape data according to our needs in the query.

Solution 4: [4]

After having to tackle this issue myself, I would like to build upon user854301's answer.

Mongoose ^4.13.8 I was able to use a function called toConstructor() which allowed me to avoid building the query multiple times when filters are applied. I know this function is available in older versions too but you'll have to check the Mongoose docs to confirm this.

The following uses Bluebird promises:

let schema = Query.find({ name: 'bloggs', age: { $gt: 30 } });

// save the query as a 'template'
let query = schema.toConstructor();

return Promise.join(

    function (total, data) {
        return { data: data, total: total }

Now the count query will return the total records it matched and the data returned will be a subset of the total records.

Please note the () around query() which constructs the query.

Solution 5: [5]

There is a library that will do all of this for you, check out mongoose-paginate-v2


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Source: Stack Overflow

Solution Credit
Solution 1 user854301
Solution 2 DhineshYes
Solution 4 oli_taz
Solution 5 Dev01